Return an array in c with pointers

Question

I am trying to make a program that makes simple things. Actually, I know how to do it but in an easier way without pointers and stuff. However, I wondered how I could do it differently (like I did below). Obviously, there is something I miss about pointers, I did the math, but still I cannot get its philosophy. Thank you!

long *read_array(int n1, int n2)
{
    int i, j;
    long a[n1][n2];
    for(i=0;i<n1;i++)
        for(j=0;j<n2;j++)
            printf("Fill the table");
    a[i][j]=GetLong();
    return a;
}

long *Min_of_Rows(int m, int n, long *a)
{
    long B[];
    int i, j;
    for(i=0;i<m;i++)
        B[i]=a[i][0];
    for(j=0;j<n;j++)
        if (a[i][j]<B[i])
            B[i]=a[i][j];

    return B;
}

void *Print_B_array (int M, long *b)
{
    int i;
    for(i=0; i<M; i++)
        printf("%ld\n",b[i]);
}

main()
{
    long *a, *b;
    int n1, n2;

    printf("give rows");
    n1=GetInteger();

    printf("give columns");
    n2=GetInteger();

    a=read_array(n1, n2);
    b=Min_of_rows(n1, n2, a);
    Print_B_Array(n1, b);

}

Please fix your indentation - it is absolutely awful to look at. — Jonathon Reinhart
– Jonathon Reinhart, Commented Nov 4, 2014 at 1:06
Also, you can't return the address of a local variable of a function and then use it after that function returns - (e.g. return a in read_array). You need to allocate space for the array with malloc. — Jonathon Reinhart
– Jonathon Reinhart, Commented Nov 4, 2014 at 1:07
long a[n1][n2]; This won't compile, since n1 and n2 are not constant. On this line (for (j = 0; j>n; j++)), n is not defined. long B[]; this is not valid C. Also, what is GetLong? — user2209008
– user2209008, Commented Nov 4, 2014 at 1:12
As I can see I cannot escape the allocation :( I will try malloc! — Doraemon
– Doraemon, Commented Nov 4, 2014 at 1:20

BLUEPIXY · Accepted Answer · 2014-11-04 01:26:33Z

1

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

long *read_array(int n1, int n2){
    int i, j;
    long a[n1][n2];

    printf("Fill the table\n");
    for(i=0;i<n1;i++){
        for(j=0;j<n2;j++){
            a[i][j]=GetLong();
        }
    }
    long *ret = malloc(sizeof(a));
    memcpy(ret, a, sizeof(a));
    return ret;
}

long *Min_of_Rows(int m, int n, long *a){
    long *B = malloc(m*sizeof(*B));
    int i, j;
    long (*A)[n] = (void*)a;

    for(i=0;i<m;i++){
        B[i]=A[i][0];
        for(j=0;j<n;j++){
            if(A[i][j]<B[i])
                B[i]=A[i][j];
        }
    }
    return B;
}

void *Print_B_Array (int M, long *b){
    int i;
    for(i=0; i<M; i++)
        printf("%ld\n", b[i]);
}

int main(void){
    long *a, *b;
    int n1, n2;

    printf("give rows:");
    n1=GetInteger();

    printf("give columns:");
    n2=GetInteger();

    a=read_array(n1, n2);
    b=Min_of_Rows(n1, n2, a);
    printf("min of rows\n");
    Print_B_Array(n1, b);
    free(a);
    free(b);
    return 0;
}

answered Nov 4, 2014 at 1:26

BLUEPIXY

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1 Comment

M.M Over a year ago

read_array uses stack space when it doesn't have to (could cause stack overflow if the dimensions are big) so I'd suggest long (*a)[n2] = malloc( n1 * sizeof *a );, then the same for loop, then return (long *)a;

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